What are the expansions of Bose functions for studying thermodynamic behavior?

[erbilsilik]

Homework Statement

To study the thermodynamic behavior of the limit $$z\rightarrow1$$ it is useful to get the expansions of $$g_{0}\left( z\right),g_{1}\left( z\right),g_{2}\left( z\right)$$

$$\alpha =-\ln z$$ which is small positive number. From, BE integral,
$$g_{1}\left( \alpha \right) =-ln\left( 1-z\right) =-ln\alpha+\dfrac {\alpha } {2}-\dfrac {\alpha ^{2}} {24}+O({\alpha ^{4}})$$
and hence

$$g_{0}\left( \alpha \right) =-\dfrac {\partial } {\partial \alpha }g_{1}\left( \alpha \right)=\dfrac {1} {\alpha }-\dfrac {1} {2}+\dfrac {\alpha } {12}O({\alpha ^{3}})$$

[Source: A.Khare, Fractional Statistics and Quantum Theory, Two Dimensional Bose Gas, p.113]

Could anyone help me to derive this expressions? I can't figure out what does it mean writing this functions in the powers of α.

Homework Equations

https://en.wikipedia.org/wiki/Polylogarithm (BE integral)

The Attempt at a Solution

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[Charles Link]

Homework Statement

To study the thermodynamic behavior of the limit $$z\rightarrow1$$ it is useful to get the expansions of $$g_{0}\left( z\right),g_{1}\left( z\right),g_{2}\left( z\right)$$

$$\alpha =-\ln z$$ which is small positive number. From, BE integral,
$$g_{1}\left( \alpha \right) =-ln\left( 1-z\right) =-ln\alpha+\dfrac {\alpha } {2}-\dfrac {\alpha ^{2}} {24}+O({\alpha ^{4}})$$
and hence

$$g_{0}\left( \alpha \right) =-\dfrac {\partial } {\partial \alpha }g_{1}\left( \alpha \right)=\dfrac {1} {\alpha }-\dfrac {1} {2}+\dfrac {\alpha } {12}O({\alpha ^{3}})$$

[Source: A.Khare, Fractional Statistics and Quantum Theory, Two Dimensional Bose Gas, p.113]

Could anyone help me to derive this expressions? I can't figure out what does it mean writing this functions in the powers of α.

Homework Equations

https://en.wikipedia.org/wiki/Polylogarithm (BE integral)

The Attempt at a Solution

[/B]

Suggestion: Write $z=exp(-\alpha)$ and you get a term $ln(exp(\alpha)-1)$ plus another term $ln(exp(\alpha))=\alpha$ In the first term (inside the $ln$ ) expand $exp(\alpha) =1+\alpha+(\alpha)^2/2 +...$ and subtract the 1. Then factor out $\alpha$ and you get $ln(\alpha)+ln(1+\alpha/2+..)$ The expansion of $ln(1+u)=u$ for small $u$.

 

[erbilsilik]

Thank you for your help Mr. @Charles Link